\(\int \frac {1}{(a+b x^2) (c+d x^2)^3} \, dx\) [650]

Optimal result

Integrand size = 19, antiderivative size = 160 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=-\frac {d x}{4 c (b c-a d) \left (c+d x^2\right )^2}-\frac {d (7 b c-3 a d) x}{8 c^2 (b c-a d)^2 \left (c+d x^2\right )}+\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} (b c-a d)^3}-\frac {\sqrt {d} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} (b c-a d)^3} \] Output:

-1/4*d*x/c/(-a*d+b*c)/(d*x^2+c)^2-1/8*d*(-3*a*d+7*b*c)*x/c^2/(-a*d+b*c)^2/ 
(d*x^2+c)+b^(5/2)*arctan(b^(1/2)*x/a^(1/2))/a^(1/2)/(-a*d+b*c)^3-1/8*d^(1/ 
2)*(3*a^2*d^2-10*a*b*c*d+15*b^2*c^2)*arctan(d^(1/2)*x/c^(1/2))/c^(5/2)/(-a 
*d+b*c)^3
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {1}{8} \left (\frac {d x \left (a d \left (5 c+3 d x^2\right )-b c \left (9 c+7 d x^2\right )\right )}{c^2 (b c-a d)^2 \left (c+d x^2\right )^2}-\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} (-b c+a d)^3}-\frac {\sqrt {d} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{5/2} (b c-a d)^3}\right ) \] Input:

Integrate[1/((a + b*x^2)*(c + d*x^2)^3),x]
 

Output:

((d*x*(a*d*(5*c + 3*d*x^2) - b*c*(9*c + 7*d*x^2)))/(c^2*(b*c - a*d)^2*(c + 
 d*x^2)^2) - (8*b^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*(-(b*c) + a* 
d)^3) - (Sqrt[d]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[d]*x)/ 
Sqrt[c]])/(c^(5/2)*(b*c - a*d)^3))/8
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {316, 402, 397, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx\)

\(\Big \downarrow \) 316

\(\displaystyle \frac {\int \frac {-3 b d x^2+4 b c-3 a d}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{4 c (b c-a d)}-\frac {d x}{4 c \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\int \frac {8 b^2 c^2-7 a b d c+3 a^2 d^2-b d (7 b c-3 a d) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 c (b c-a d)}-\frac {d x (7 b c-3 a d)}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {8 b^3 c^2 \int \frac {1}{b x^2+a}dx}{b c-a d}-\frac {d \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \int \frac {1}{d x^2+c}dx}{b c-a d}}{2 c (b c-a d)}-\frac {d x (7 b c-3 a d)}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {8 b^{5/2} c^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} (b c-a d)}-\frac {\sqrt {d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)}}{2 c (b c-a d)}-\frac {d x (7 b c-3 a d)}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (c+d x^2\right )^2 (b c-a d)}\)

Input:

Int[1/((a + b*x^2)*(c + d*x^2)^3),x]
 

Output:

-1/4*(d*x)/(c*(b*c - a*d)*(c + d*x^2)^2) + (-1/2*(d*(7*b*c - 3*a*d)*x)/(c* 
(b*c - a*d)*(c + d*x^2)) + ((8*b^(5/2)*c^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(S 
qrt[a]*(b*c - a*d)) - (Sqrt[d]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTa 
n[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)))/(2*c*(b*c - a*d)))/(4*c*(b* 
c - a*d))
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.99

method	result	size
default	\(-\frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\left (a d -b c \right )^{3} \sqrt {a b}}+\frac {d \left (\frac {\frac {d \left (3 a^{2} d^{2}-10 a b c d +7 b^{2} c^{2}\right ) x^{3}}{8 c^{2}}+\frac {\left (5 a^{2} d^{2}-14 a b c d +9 b^{2} c^{2}\right ) x}{8 c}}{\left (x^{2} d +c \right )^{2}}+\frac {\left (3 a^{2} d^{2}-10 a b c d +15 b^{2} c^{2}\right ) \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{8 c^{2} \sqrt {c d}}\right )}{\left (a d -b c \right )^{3}}\)	\(158\)
risch	\(\text {Expression too large to display}\)	\(2285\)

Input:

int(1/(b*x^2+a)/(d*x^2+c)^3,x,method=_RETURNVERBOSE)
 

Output:

-b^3/(a*d-b*c)^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))+d/(a*d-b*c)^3*((1/8*d 
*(3*a^2*d^2-10*a*b*c*d+7*b^2*c^2)/c^2*x^3+1/8*(5*a^2*d^2-14*a*b*c*d+9*b^2* 
c^2)/c*x)/(d*x^2+c)^2+1/8*(3*a^2*d^2-10*a*b*c*d+15*b^2*c^2)/c^2/(c*d)^(1/2 
)*arctan(x*d/(c*d)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (138) = 276\).

Time = 0.60 (sec) , antiderivative size = 1585, normalized size of antiderivative = 9.91 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="fricas")
 

Output:

[-1/16*(2*(7*b^2*c^2*d^2 - 10*a*b*c*d^3 + 3*a^2*d^4)*x^3 + 8*(b^2*c^2*d^2* 
x^4 + 2*b^2*c^3*d*x^2 + b^2*c^4)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) 
- a)/(b*x^2 + a)) + (15*b^2*c^4 - 10*a*b*c^3*d + 3*a^2*c^2*d^2 + (15*b^2*c 
^2*d^2 - 10*a*b*c*d^3 + 3*a^2*d^4)*x^4 + 2*(15*b^2*c^3*d - 10*a*b*c^2*d^2 
+ 3*a^2*c*d^3)*x^2)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + 
 c)) + 2*(9*b^2*c^3*d - 14*a*b*c^2*d^2 + 5*a^2*c*d^3)*x)/(b^3*c^7 - 3*a*b^ 
2*c^6*d + 3*a^2*b*c^5*d^2 - a^3*c^4*d^3 + (b^3*c^5*d^2 - 3*a*b^2*c^4*d^3 + 
 3*a^2*b*c^3*d^4 - a^3*c^2*d^5)*x^4 + 2*(b^3*c^6*d - 3*a*b^2*c^5*d^2 + 3*a 
^2*b*c^4*d^3 - a^3*c^3*d^4)*x^2), -1/8*((7*b^2*c^2*d^2 - 10*a*b*c*d^3 + 3* 
a^2*d^4)*x^3 + (15*b^2*c^4 - 10*a*b*c^3*d + 3*a^2*c^2*d^2 + (15*b^2*c^2*d^ 
2 - 10*a*b*c*d^3 + 3*a^2*d^4)*x^4 + 2*(15*b^2*c^3*d - 10*a*b*c^2*d^2 + 3*a 
^2*c*d^3)*x^2)*sqrt(d/c)*arctan(x*sqrt(d/c)) + 4*(b^2*c^2*d^2*x^4 + 2*b^2* 
c^3*d*x^2 + b^2*c^4)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 
+ a)) + (9*b^2*c^3*d - 14*a*b*c^2*d^2 + 5*a^2*c*d^3)*x)/(b^3*c^7 - 3*a*b^2 
*c^6*d + 3*a^2*b*c^5*d^2 - a^3*c^4*d^3 + (b^3*c^5*d^2 - 3*a*b^2*c^4*d^3 + 
3*a^2*b*c^3*d^4 - a^3*c^2*d^5)*x^4 + 2*(b^3*c^6*d - 3*a*b^2*c^5*d^2 + 3*a^ 
2*b*c^4*d^3 - a^3*c^3*d^4)*x^2), -1/16*(2*(7*b^2*c^2*d^2 - 10*a*b*c*d^3 + 
3*a^2*d^4)*x^3 - 16*(b^2*c^2*d^2*x^4 + 2*b^2*c^3*d*x^2 + b^2*c^4)*sqrt(b/a 
)*arctan(x*sqrt(b/a)) + (15*b^2*c^4 - 10*a*b*c^3*d + 3*a^2*c^2*d^2 + (15*b 
^2*c^2*d^2 - 10*a*b*c*d^3 + 3*a^2*d^4)*x^4 + 2*(15*b^2*c^3*d - 10*a*b*c...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x**2+a)/(d*x**2+c)**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (138) = 276\).

Time = 0.14 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.73 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a b}} - \frac {{\left (15 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} \sqrt {c d}} - \frac {{\left (7 \, b c d^{2} - 3 \, a d^{3}\right )} x^{3} + {\left (9 \, b c^{2} d - 5 \, a c d^{2}\right )} x}{8 \, {\left (b^{2} c^{6} - 2 \, a b c^{5} d + a^{2} c^{4} d^{2} + {\left (b^{2} c^{4} d^{2} - 2 \, a b c^{3} d^{3} + a^{2} c^{2} d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{5} d - 2 \, a b c^{4} d^{2} + a^{2} c^{3} d^{3}\right )} x^{2}\right )}} \] Input:

integrate(1/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="maxima")
 

Output:

b^3*arctan(b*x/sqrt(a*b))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3* 
d^3)*sqrt(a*b)) - 1/8*(15*b^2*c^2*d - 10*a*b*c*d^2 + 3*a^2*d^3)*arctan(d*x 
/sqrt(c*d))/((b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 - a^3*c^2*d^3)*sqr 
t(c*d)) - 1/8*((7*b*c*d^2 - 3*a*d^3)*x^3 + (9*b*c^2*d - 5*a*c*d^2)*x)/(b^2 
*c^6 - 2*a*b*c^5*d + a^2*c^4*d^2 + (b^2*c^4*d^2 - 2*a*b*c^3*d^3 + a^2*c^2* 
d^4)*x^4 + 2*(b^2*c^5*d - 2*a*b*c^4*d^2 + a^2*c^3*d^3)*x^2)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a b}} - \frac {{\left (15 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} \sqrt {c d}} - \frac {7 \, b c d^{2} x^{3} - 3 \, a d^{3} x^{3} + 9 \, b c^{2} d x - 5 \, a c d^{2} x}{8 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} {\left (d x^{2} + c\right )}^{2}} \] Input:

integrate(1/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="giac")
 

Output:

b^3*arctan(b*x/sqrt(a*b))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3* 
d^3)*sqrt(a*b)) - 1/8*(15*b^2*c^2*d - 10*a*b*c*d^2 + 3*a^2*d^3)*arctan(d*x 
/sqrt(c*d))/((b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 - a^3*c^2*d^3)*sqr 
t(c*d)) - 1/8*(7*b*c*d^2*x^3 - 3*a*d^3*x^3 + 9*b*c^2*d*x - 5*a*c*d^2*x)/(( 
b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*(d*x^2 + c)^2)
 

Mupad [B] (verification not implemented)

Time = 2.51 (sec) , antiderivative size = 6033, normalized size of antiderivative = 37.71 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

int(1/((a + b*x^2)*(c + d*x^2)^3),x)
 

Output:

((x^3*(3*a*d^3 - 7*b*c*d^2))/(8*c^2*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (x* 
(5*a*d^2 - 9*b*c*d))/(8*c*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/(c^2 + d^2*x^4 
 + 2*c*d*x^2) - (atan((((-a*b^5)^(1/2)*((x*(9*a^4*b^3*d^7 + 289*b^7*c^4*d^ 
3 - 300*a*b^6*c^3*d^4 - 60*a^3*b^4*c*d^6 + 190*a^2*b^5*c^2*d^5))/(32*(b^4* 
c^8 + a^4*c^4*d^4 - 4*a^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^3*c^7*d)) 
- ((-a*b^5)^(1/2)*((256*b^10*c^10*d^2 - 1760*a*b^9*c^9*d^3 + 5280*a^2*b^8* 
c^8*d^4 - 9056*a^3*b^7*c^7*d^5 + 9760*a^4*b^6*c^6*d^6 - 6816*a^5*b^5*c^5*d 
^7 + 3040*a^6*b^4*c^4*d^8 - 800*a^7*b^3*c^3*d^9 + 96*a^8*b^2*c^2*d^10)/(64 
*(b^6*c^10 + a^6*c^4*d^6 - 6*a^5*b*c^5*d^5 + 15*a^2*b^4*c^8*d^2 - 20*a^3*b 
^3*c^7*d^3 + 15*a^4*b^2*c^6*d^4 - 6*a*b^5*c^9*d)) - (x*(-a*b^5)^(1/2)*(256 
*b^9*c^11*d^2 - 1280*a*b^8*c^10*d^3 + 2304*a^2*b^7*c^9*d^4 - 1280*a^3*b^6* 
c^8*d^5 - 1280*a^4*b^5*c^7*d^6 + 2304*a^5*b^4*c^6*d^7 - 1280*a^6*b^3*c^5*d 
^8 + 256*a^7*b^2*c^4*d^9))/(64*(a^4*d^3 - a*b^3*c^3 + 3*a^2*b^2*c^2*d - 3* 
a^3*b*c*d^2)*(b^4*c^8 + a^4*c^4*d^4 - 4*a^3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 
- 4*a*b^3*c^7*d))))/(2*(a^4*d^3 - a*b^3*c^3 + 3*a^2*b^2*c^2*d - 3*a^3*b*c* 
d^2)))*1i)/(2*(a^4*d^3 - a*b^3*c^3 + 3*a^2*b^2*c^2*d - 3*a^3*b*c*d^2)) + ( 
(-a*b^5)^(1/2)*((x*(9*a^4*b^3*d^7 + 289*b^7*c^4*d^3 - 300*a*b^6*c^3*d^4 - 
60*a^3*b^4*c*d^6 + 190*a^2*b^5*c^2*d^5))/(32*(b^4*c^8 + a^4*c^4*d^4 - 4*a^ 
3*b*c^5*d^3 + 6*a^2*b^2*c^6*d^2 - 4*a*b^3*c^7*d)) + ((-a*b^5)^(1/2)*((256* 
b^10*c^10*d^2 - 1760*a*b^9*c^9*d^3 + 5280*a^2*b^8*c^8*d^4 - 9056*a^3*b^...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 571, normalized size of antiderivative = 3.57 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {-8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c^{5}-16 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c^{4} d \,x^{2}-8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c^{3} d^{2} x^{4}+3 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{3} c^{2} d^{2}+6 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{3} c \,d^{3} x^{2}+3 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{3} d^{4} x^{4}-10 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} b \,c^{3} d -20 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} b \,c^{2} d^{2} x^{2}-10 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} b c \,d^{3} x^{4}+15 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a \,b^{2} c^{4}+30 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a \,b^{2} c^{3} d \,x^{2}+15 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a \,b^{2} c^{2} d^{2} x^{4}+5 a^{3} c^{2} d^{3} x +3 a^{3} c \,d^{4} x^{3}-14 a^{2} b \,c^{3} d^{2} x -10 a^{2} b \,c^{2} d^{3} x^{3}+9 a \,b^{2} c^{4} d x +7 a \,b^{2} c^{3} d^{2} x^{3}}{8 a \,c^{3} \left (a^{3} d^{5} x^{4}-3 a^{2} b c \,d^{4} x^{4}+3 a \,b^{2} c^{2} d^{3} x^{4}-b^{3} c^{3} d^{2} x^{4}+2 a^{3} c \,d^{4} x^{2}-6 a^{2} b \,c^{2} d^{3} x^{2}+6 a \,b^{2} c^{3} d^{2} x^{2}-2 b^{3} c^{4} d \,x^{2}+a^{3} c^{2} d^{3}-3 a^{2} b \,c^{3} d^{2}+3 a \,b^{2} c^{4} d -b^{3} c^{5}\right )} \] Input:

int(1/(b*x^2+a)/(d*x^2+c)^3,x)
 

Output:

( - 8*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**2*c**5 - 16*sqrt(b) 
*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**2*c**4*d*x**2 - 8*sqrt(b)*sqrt(a 
)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**2*c**3*d**2*x**4 + 3*sqrt(d)*sqrt(c)*at 
an((d*x)/(sqrt(d)*sqrt(c)))*a**3*c**2*d**2 + 6*sqrt(d)*sqrt(c)*atan((d*x)/ 
(sqrt(d)*sqrt(c)))*a**3*c*d**3*x**2 + 3*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d 
)*sqrt(c)))*a**3*d**4*x**4 - 10*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c 
)))*a**2*b*c**3*d - 20*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2* 
b*c**2*d**2*x**2 - 10*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*b 
*c*d**3*x**4 + 15*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b**2*c** 
4 + 30*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b**2*c**3*d*x**2 + 
15*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b**2*c**2*d**2*x**4 + 5 
*a**3*c**2*d**3*x + 3*a**3*c*d**4*x**3 - 14*a**2*b*c**3*d**2*x - 10*a**2*b 
*c**2*d**3*x**3 + 9*a*b**2*c**4*d*x + 7*a*b**2*c**3*d**2*x**3)/(8*a*c**3*( 
a**3*c**2*d**3 + 2*a**3*c*d**4*x**2 + a**3*d**5*x**4 - 3*a**2*b*c**3*d**2 
- 6*a**2*b*c**2*d**3*x**2 - 3*a**2*b*c*d**4*x**4 + 3*a*b**2*c**4*d + 6*a*b 
**2*c**3*d**2*x**2 + 3*a*b**2*c**2*d**3*x**4 - b**3*c**5 - 2*b**3*c**4*d*x 
**2 - b**3*c**3*d**2*x**4))